1. We check the hybridization of the central atom in each species using steric number:

$\text{Steric number} = \text{number of }\sigma\text{-bonds} + \text{number of lone pairs on central atom}$

If steric number $=4$, then hybridization is $\mathrm{sp}^3$.

2. Analyze each species:

(i) $\mathrm{NO_3^-}$

Central atom: $\mathrm{N}$

• Nitrogen is bonded to 3 oxygens.
• It has 3 $\sigma$-bonds and 0 lone pairs.

So, steric number:

$3 + 0 = 3$

Hence hybridization is:

$\mathrm{sp}^2$

So, not $\mathrm{sp}^3$.

(ii) $\mathrm{BCl_3}$

Central atom: $\mathrm{B}$

• Boron forms 3 $\sigma$-bonds with 3 chlorine atoms.
• It has 0 lone pairs.

Steric number:

$3 + 0 = 3$

Hence hybridization is:

$\mathrm{sp}^2$

So, not $\mathrm{sp}^3$.

(iii) $\mathrm{ClO_2^-}$

Central atom: $\mathrm{Cl}$

Total valence electrons:

$7 + 2(6) + 1 = 20$

Structure has 2 Cl–O bonds and after completing octets, chlorine has 2 lone pairs. Thus around chlorine:

• 2 $\sigma$-bonds
• 2 lone pairs

Steric number:

$2 + 2 = 4$

Hence hybridization is:

$\mathrm{sp}^3$

So, this is $\mathrm{sp}^3$.

(iv) $\mathrm{ClO_3^-}$

Central atom: $\mathrm{Cl}$

Total valence electrons:

$7 + 3(6) + 1 = 26$

Chlorine is bonded to 3 oxygens and has 1 lone pair on the central atom. Thus around chlorine:

• 3 $\sigma$-bonds
• 1 lone pair

Steric number:

$3 + 1 = 4$

Hence hybridization is:

$\mathrm{sp}^3$

So, this is $\mathrm{sp}^3$.

3. Count species with central atom in $\mathrm{sp}^3$ hybridization:

• $\mathrm{NO_3^-}$ : No
• $\mathrm{BCl_3}$ : No
• $\mathrm{ClO_2^-}$ : Yes
• $\mathrm{ClO_3^-}$ : Yes

Therefore, number of such species is:

$2$

So the correct option is A.